Hey I'm not sure how you solve this. Calculate to the nearest degree the values of x when sin x = 0.2, for -180degrees < x < 180degrees.

Apr 23rd 2006, 01:47 PM

CaptainBlack

Quote:

Originally Posted by digideens

Hey I'm not sure how you solve this. Calculate to the nearest degree the values of x when sin x = 0.2, for -180degrees < x < 180degrees.

It depends on what you mean by calculate. I would normaly use the inverse
sin function on a calculator to do this. It can be done using Newton Raphson
or a number of other techniques.

There is also the problem that there are two solutions in the specified range.

RonL

Apr 23rd 2006, 01:48 PM

ThePerfectHacker

Quote:

Originally Posted by digideens

Hey I'm not sure how you solve this. Calculate to the nearest degree the values of x when sin x = 0.2, for -180degrees < x < 180degrees.

You need to use a function on your calculator called the arc-sin or sometimes written as $\displaystyle \sin^{-1}$ it tell you the inverse operation.
Thus, for
$\displaystyle \sin x=.2$
You find, (have calculator in degree mode).
$\displaystyle x=\sin^{-1}(.2)\approx 11.54^o$
But that is still not it. Because the sine is also positive in the second Quadrant. In the second quadrant the value is,
$\displaystyle 180^o-11.54^o\approx 168.46^o$